Bouncing Back to P

Relevant wiki: Analyzing inelastic collisions

Let initial velocity of ball be u projected at an angle $\theta$ .

$v_{x} = u\cos\theta$ => Velocity of ball just before first collision.

Therefore, velocity of ball just after first collision = $eu\cos\theta$

Let PA = $d_{2}$ and AB= $d_{1}$ [ $d_{1} + d_{2}$ = PB] -------------- (1)

We know that PB = $(u^{2}\sin\theta\cos\theta)/g$ ------------------- (2)

Now $d_{1}$ = $(eu^{2}\sin\theta\cos\theta)/g$ ----------------- (3)

After second collision, ball has $v_{x} = eu\cos\theta$ and $v_{y} = eu\sin\theta$

Therefore $d_{2}$ = $(2e^{2}u^{2}\sin\theta\cos\theta)/g$ ------------------ (4)

From eqns. (1),(2),(3) and (4) ,

$(2e^{2}+e)(u^{2}\sin\theta\cos\theta/g) = (u^{2}\sin\theta\cos\theta/g)$

=>We get, $2e^{2}+e-1 = 0$

Solving the equation we get

e = 1/2 and e = -1

But e > 0 .

So e = 1/2